Found Identity – Order of Group Elements

May 18, 2009 14:34 by scibuff

Another one in my series of Yahoo Answers this time using elementary properties of Groups.

Problem: Let a,b\ \epsilon\ G be elements of group such that (ab)^{k}=e where e is the identity element of G. Prove that (ba)^{k}=e

Solution: The problem is indeed trivial if G is abelian, i.e. ab=ba for all a,b\ \epsilon\ G. One must be careful not to assume such properties when dealing with constructs in abstract algebra. Nevertheless, the solution below is quite simple and uses only associativity of Groups, i.e. (ab)c=a(bc) for all a,b,c\ \epsilon\ G.

Lemma: For any a,b\ \epsilon\ G there is a unique x\ \epsilon\ G such that ax=b.

Proof: The lemma can be proven in two steps. First, show that such x exists, and second, show that it must be unique.

Let x=a^{-1}b. Since a,b\ \epsilon\ G and G is a group, it follows from the existence of  inverses that a^{-1}\ \epsilon\ G and from group closure a^{-1}b\ \epsilon\ G. Thus, x\ \epsilon\ G (and the first step is done).

Let x\ \epsilon\ G be such that ax=b, take e to be the identity element of G. Then


Thus it follows that if ax=b then it must be that x=a^{-1}b, finishing the second step, hence proving out lemma.

Now we go back to our proof. Suppose


is the identity in G, i.e.

\overset{k\ times}{\overbrace{(ab)(ab)\cdots (ab)}}=e\ (1)

Using group associativity we get

a\overset{k-1\ times}{\overbrace{(ba)(ba)\cdots (ba)}}b=a(ba)^{k-1}b=e\ (2)

Multiplying(2) by a (on right) we get


a(ba)^{k}=a\ (3)

Now multiplying (2) by b (on left) we get


(ba)^{k}b=b\ (4)

Now from (3) we have

ba(ba)^{k}=ba\ (5)

and from (4)

(ba)^{k}ba=ba\ (6)

Combining (5) and (6)

ba(ba)^{k}=(ba)^{k}ba=ba\ (7)

Since e is the identity element of G, by definition

(ba)e=e(ba)=ba\ (8)

Finally, using our Lemma and (7) with (8) it follows that (ba)^{k}=e